杭电ACM试题答案(共4页)

1、【杭电ACM1000】A + B ProblemProblem DescriptionCalculate A + B.InputEach line will contain two integers A and B. Process to end of file.OutputFor each case, output A + B in one line.Sample Input1 1Sample Output2# include <stdio.h>int main()int a, b;while(scanf("%d%d", &a, &b)!=EO

2、F)printf("%dn", a+b);return 0; 【杭电ACM1001】Sum ProblemProblem DescriptionHey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + . + n.InputThe input will consist of a series of integers n, one integer per line.OutputFor

3、each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.Sample Input1 100Sample Output1 5050# include <stdio.h>int main()int n, i, sum = 0;while(scanf("%d", &n)!=EOF)for(i=1; i<=n; +i)sum = sum + i

4、;printf("%dnn", sum);sum = 0;return 0; 【杭电ACM1002】A + B Problem IIProblem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.InputThe first line of the input contains an integer T(1<=T<=20) which means the number of

5、test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.OutputFor each test case, you should output t

6、wo lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.Sample Input2 1 2 11223344556677889

7、9 998877665544332211Sample OutputCase 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110#include<stdio.h>#include<string.h>int shu(char a) return (a-'0');int main() char a1000,b1000; int num1001; int n,i,j=1,al,bl,k,t;scanf("%d",&n);

8、while(n-) getchar(); if(j!=1)printf("n"); scanf("%s",a); al=strlen(a); scanf("%s",b); bl=strlen(b); k=(al>bl)?al:bl; for(i=0;i<=k;i+) numi=0; t=k; for(k;al>0&&bl>0;k-) numk+=shu(a-al)+shu(b-bl); if(numk/10) numk-1+; numk%=10; while(al>0) numk-+=shu(a-al); if(numk+1/10) numk+; numk+1%=10; while(bl>0) numk-+=shu(b-bl); if(numk+1/10) numk+; numk+1%=10; printf("Case %d:n",j+); printf(&quo